考慮Br和H為活性中間體,在其量較多時(shí)一旦有新的生成馬上就消耗掉,濃度接近不變（穩(wěn)態(tài)近似）.第一步反應(yīng)應(yīng)該是可逆的,否則無法推導(dǎo).各步的速率常數(shù)分別為k1、k-1（第一步的逆反應(yīng)）、k2、k3.
則有
①dc(Br)/dt=2k1c(Br2)-k-1[c(Br)]^2-k2c(Br)c(H)+k3c(H)c(Br2)=0
②dc(H)/dt=k2c(Br)c(H2)-k3c(H)c(Br2)=0
由此可解得c(Br)=[2k1c(Br2)/k-1]^0.5,c(H)=k2[2k1c(Br2)/k-1]^0.5·c(H2)/[k3c(Br2)]
由總反應(yīng)速率的定義,可知
r=dc(HBr)/(2dt)
=k2c(Br)c(H2)+k3c(H)c(Br2)
=k2[2k1c(Br2)/k-1]^0.5·c(H2)
若用dc(HBr)/dt表示,則再乘2即可.